# Monatomic lattice energy density and Kubo formula

Let $$\mathcal{L}$$ be the lattice at temperature close to zero so that we may assume that the system is in the vacuum state. The Hamiltonian of the system is given by:

$$H=\frac{1}{2m}\sum_{x\in \mathcal{L}}p(x)^2 + \sum_{x \in \mathcal{L}} \frac{\kappa}{2}(2q(x)^2-q(x)q(x+b)-q(x)q(x-b)).$$

The energy density $$u(x)$$ is defined by $$H = \sum_{x \in \mathcal{L}}u(x)$$.

I have to rewrite the energy density in terms of creation and annihilation operators. Notice that the frequencies are given by:

$$\omega_k^2 = 4\frac{\kappa}{m}\sin^2(kb/2)$$

where $$b$$ is the lattice constant.

Recall that the position and momentum operators are defined by:

$$q(x) = \sum_{k} \sqrt{\frac{\hbar}{2Nm\omega_k}}(a_k e^{ikx}+a_k^\dagger e^{-ikx})$$

$$p(x)=-i\sum_k \sqrt{\hbar m \omega_k/2N}(a_k e^{ikx}-a_k^\dagger e^{-ikx})$$

where we assumed $$N$$ particles with same mass $$m$$ and the sum is over the 1st Brillouin zone.

I determined the energy density in terms of these operators:

$$u(x)= -\frac{\hbar}{4N} \sum_{k,l} \sqrt{\omega_k \omega_l} (a_k a_l e^{i(k+l)x} + a_k^\dagger a_l^\dagger e^{-i(k+l)x} - a_k a_l^\dagger e^{i(k-l)x}-a_k^\dagger a_l e^{i(l-k)x}) + \frac{\kappa}{2} \frac{\hbar}{2Nm} \sum_{k,l} 4\sin^2(lb/2) (a_k a_l e^{i(k+l)x} + a_k^\dagger a_l^\dagger e^{-i(k+l)x} + a_k^\dagger a_l e^{i(l-k)x} + a_k a_l^\dagger e^{i(k-l)x})\frac{1}{\sqrt{\omega_k \omega_l}}$$

I used that $$2-2cos(lb) = 4\sin^2(lb/2)$$.

Using commutator rule $$[a_k,a_l^\dagger] = \delta_{kl}$$ we can write this in normal order denoted $$u(x) = :u(x): + const.$$ where normal order means only terms where annihilation operators are right and creation operators are left.

In this way I could write the previous expression as

$$u(x)= -\frac{\hbar}{4N} \sum_{k,l} \sqrt{\omega_k \omega_l} (a_k a_l e^{i(k+l)x} + a_k^\dagger a_l^\dagger e^{-i(k+l)x} - a_l^\dagger a_k e^{i(k-l)x}-a_k^\dagger a_l e^{i(l-k)x}) + \frac{\kappa}{2} \frac{\hbar}{2Nm} \sum_{k,l} 4\sin^2(lb/2) (a_k a_l e^{i(k+l)x} + a_k^\dagger a_l^\dagger e^{-i(k+l)x} + a_k^\dagger a_l e^{i(l-k)x} - a_l^\dagger a_k e^{i(k-l)x})\frac{1}{\sqrt{\omega_k \omega_l}}+ \frac{\hbar}{2N} \sum_{k} \omega_k.$$

The idea is that we perturb the atom at $$x=0$$ at time $$t=0$$ such that the mass increases from $$m$$ to $$M$$ (where $$M-m \ll m$$):

$$\frac{p(0)^2}{2M} = \frac{p(0)^2}{2m(\frac{M-m}{m}+1)} \approx \frac{p(0)^2}{2m} - \frac{p(0)^2}{2m^2}(M-m),$$

therfore the perturbation of $$H$$ is given by $$\delta H (t) = -\frac{p(0)^2}{2m^2}(M-m)\theta(t)$$ where $$\theta(t)$$ is the step function (equal to zero for $$t<0$$ and one for $$t>0$$). From this I could use the Kubo formula for $$:u(x,t):$$ using the Heisenberg representation of $$a_k \rightarrow a_k(t) = e^{-i\omega_k t}a_k$$ for all the terms in $$u(x)$$.

$$\langle :u(x,t): \rangle = -\frac{i}{\hbar}\int_0^t \langle [:u(x,t):,p^2(0,t^\prime)]\rangle dt^\prime$$

where I ignored the prefactor in $$\delta H$$ for simplicity. I took the vacuum expectation value as was asked in the problem and I ended up with the following propagator (integrand of the given integral, taking $$t^\prime = 0$$):

$$D^R(x,t) \sim \int (-\omega_k \omega_l + \omega_l^2)\sin((k+l)x-(\omega_k + \omega_l)t)\theta(t) dk dl$$

I have to show or argue that this vanishes for $$x for $$x\gg b$$ where $$v_s$$ is the speed of sound (defined as $$\omega_k^\prime$$ with $$k$$ close to zero). Any ideas how I can show this from this result or did I make any mistakes in my calculation?

I know that I should use an integral of the form for $$x\gg b$$:

$$\int_k dk \frac{1}{\omega_k} \sin(kx-\omega_k t) = -\frac{\pi}{v_s}\theta(x-v_s t),$$

but I cannot apply this in this case as it seems. What goes wrong?

I see no mistakes in your formulas. I've obtained the same expression for the propagator $$D^R(x,t)$$. I think it is impossible to show, that $$D^R(x,t)$$ is exactly zero if $$|x| > v_s t$$. The equality $$D^R(x,t) = 0$$ is valid asymptotically when $$|x| \gg b$$ and $$|x|> v_st$$. To demonstrate this fact, let's rewrite the propagator in the following form $$D^R(x,t) \sim \mbox{Im} \int F(k,l)\, e^{i\,x\,(S(k)\, + \,S(l))}\, \theta(t)\, dkdl,\quad (1)$$ where $$S(k) = k - \omega_k\, t/x,$$ $$F(k,l)$$ is of obvious form and doesn't depend on $$x$$. Due to $$k \sim 1/b$$, $$e^{ixS(k)}$$ is strongly oscillating function when $$|x| \gg b$$. Asymptotics of integrals with a strongly oscillating exponential function whose phase is real can be found by the Method of stationary phase. According to the Method, the main contribution to the integral (1) is due to the neighborhoods of the phase's stationary points. If the phase has no stationary points, then the integral is small as $$O(1/x^\infty)$$ when $$x\to \infty$$. Stationary point equations for the integral (1) have the following form: $$\frac{\partial}{\partial k}(S(k)+S(l)) = S'(k) = 0,\quad \frac{\partial}{\partial l}(S(k)+S(l)) = S'(l) = 0.$$ We have: $$S'(k) = 1 - \omega_k\,' t/x,$$ and $$\omega_k'$$ achieves its maximum absolute value at $$k = 0$$. Hence, the stationary point existence condition is $$|x| \leq|\omega_k\,'|_{k=0} t = v_s t.$$ If this condition is not satisfied, the propagator (1) has very small, almost zero, value.